Detemine the equation in standard form for the quadratic equation

Question

x-intercepts -2 (+or-) square root of 5, containing the point (-4, 5)

Answer 1

x = -2+sqr5, -2-sqr5

factors of the quadratic polynomial are

x+2+sqr5 and x+2-sqr5

multiply the factors together to get

(x+2)^2 - 5 = x^2 +4x +4 -5 = x^2 +4x -1

the point (-4,5) means x=-4 and y=f(x)= 5

f(x) = a(x^2 +4x -1) solve for a by plugging in x=-4 and y=5

5 =a(16-16-1) = -a

a = -5

the quadratic function is

y=-5(x^2 +4x -1)

y = -5x^2 -20x +5

is in standard y=ax^2 +bx + c form where you can apply the quadratic formula x=-b/2a + or- (1/2)sqr(b^2-4ac)

x = 20/-10 + or - (1/10)sqr(400+100) -2 + or -sqr5